Hi,
I increased the size of the interval by 66% in these two cases and it only affects the answer in the 15th digit. I
think that is a sound reason to believe it is very close to being right.
Rich
--------------------------------------------------
From: "dlakelan" <dlakelan at street-artists.org>
Sent: Thursday, March 18, 2010 9:36 PM
To: <maxima at math.utexas.edu>
Subject: Re: [Maxima] numerical integration
> On 03/18/2010 03:15 PM, Richard Hennessy wrote:
>> I got it this way.
>>
>>
>> bromberg(x^x^x, x, 599975/100000b0, 6b0),brombergit=15,fpprec:2000;
>> -> 1.1026649993619407788421378964b36300
>> bromberg(x^x^x, x, 599985/100000b0, 6b0),brombergit=15,fpprec:2000;
>>> - 1.1026649993619421259432920899b36300
>>
>> When you are this close to 6 the area under the curve between 5 and
>> 5.99985 is irrelevant since it does not affect the answer (at least not
>> the first 8-10 digits.)
>
> Yes, this is similar to Raymond Toy's solution. I get a similar answer with trapezoid rule and an estimate of the
> truncation error that assumes similitude between the truncated part and the calculated part.
>
> Is there a solution in terms of an asymptotic series? we assume integral(f(x),x,5,6) ~ integral(f(x),x,6-eps,6),
> expand the approximation in terms of the small parameter eps, and then truncate the expansion at some small number of
> terms (perhaps 2 or 3) and get some kind of decent approximate result?
>
> I tried to do this via integration by parts taking
>
> x^x^x = x * x^(x^x-1)
>
> take x dx = dv and x^(x^x-1) = u, plug into uv - int(v du), iterate again, and then chop off the remaining integral...
> but it got me nowhere, the resulting series explodes.
>
> I still feel like there should be a "clever" way to get good accuracy using some kind of asymptotic method. This type
> of extremely fast growing integral is a very special sort of thing.
>
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